Labware - MA35 Multivariable Calculus - Single Variable Calculus



Decomposition of Acceleration (Page: 1 | 2 )


We can approach curvature in terms of the circular images of the vectors T(t) and U(t). Since T(t) is a unit vector, it lies on the unit circle. So, there exists a function θ(t) such that T(t)=(cos(θ(t)), sin(θ(t))). Similarly, by the definition of normal vector, we have U(t) = (-sin(θ(t)), cos(θ(t))). Then,

T'(t) = (-sin(θ(t))θ'(t),cos(θ(t))θ'(t)) = U(t)θ'(t)

Entering this identity into the expression for acceleration yeilds X''(t) = s''(t)T(t) + s'(t)θ'(t)U(t). Earlier we defined the normal component of the accleration vector to be s'(t)2κg(t). Therefore, s'(t)θ'(t) = s'(t)2κg(t), or equivalently dθ/dt = (ds/dt)κg(t). Using an arclength parameterization, we then have

κg(s) = θ'(s)

This provides a definition of geodesic curvature in terms of the motion of the tangent and normal vectors on the unit circle.